Divide the following complex numbers. $ \dfrac{-9+21i}{3-3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+3i}$ $ \dfrac{-9+21i}{3-3i} = \dfrac{-9+21i}{3-3i} \cdot \dfrac{{3+3i}}{{3+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-9+21i) \cdot (3+3i)} {(3-3i) \cdot (3+3i)} = \dfrac{(-9+21i) \cdot (3+3i)} {3^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-9+21i) \cdot (3+3i)} {(3)^2 - (-3i)^2} = $ $ \dfrac{(-9+21i) \cdot (3+3i)} {9 + 9} = $ $ \dfrac{(-9+21i) \cdot (3+3i)} {18} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-9+21i}) \cdot ({3+3i})} {18} = $ $ \dfrac{{-9} \cdot {3} + {21} \cdot {3 i} + {-9} \cdot {3 i} + {21} \cdot {3 i^2}} {18} $ Evaluate each product of two numbers. $ \dfrac{-27 + 63i - 27i + 63 i^2} {18} $ Finally, simplify the fraction. $ \dfrac{-27 + 63i - 27i - 63} {18} = \dfrac{-90 + 36i} {18} = -5+2i $